A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{3}$ $(pi)/3$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 9, what is the area of the triangle?

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{3}$ $(pi)/3$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 9, what is the area of the triangle?

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Answer 1 · 2015-12-28T02:09:50

17.537

Explanation:

As the sum of all the angles of a triangle is $180^{0}$ $180^0$ or $π$ $pi$ radians
So, from your sum it can be seen that the third angle between A and C $= π - \frac{π}{3} - \frac{π}{6} = \frac{π}{2}$ $=pi-pi/3-pi/6=pi/2$
Thus making the side B, which is opposite to the $90^{0}$ $90^0$ angle hypotenuse of the triangle.
So, the area of the triangle would be $\frac{1}{2} \cdot b a s e \cdot h e i g h t = \frac{1}{2} \cdot A \cdot C$ $1/2*base*height=1/2*A*C$
Given that side $B = 9$ $B=9$ , we can find the other two side by triangle geometry
$A = B \cdot cos (\frac{π}{3}) = 9 \cdot cos (\frac{π}{3}) = 4.5$ $A=B*cos(pi/3)=9*cos(pi/3)=4.5$
$C = B \cdot sin (\frac{π}{6}) = 9 \cdot sin (\frac{π}{6}) = 7.794$ $C=B*sin(pi/6)=9*sin(pi/6)=7.794$
Hence, $A r e a = \frac{1}{2} \cdot A \cdot C = \frac{1}{2} \cdot 4.5 \cdot 7.794 = 17.537$ $Area=1/2*A*C=1/2*4.5*7.794=17.537$

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{3}$ $(pi)/3$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 9, what is the area of the triangle?

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{3}$ $(pi)/3$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 9, what is the area of the triangle?